TITLE: Isomorphic quotient groups $\frac{G}{H} \cong \frac{G}{K}$ imply $H \cong K$?
QUESTION [12 upvotes]: I know that given a group $G$ and two normal subgroups $H,K \subset G$ then it is not true that:
"if $H \cong K$ then $ \frac{G}{H} \cong \frac{G}{K} $ (the counterexample is quite easy with products of cyclic groups) "
My question is: Is the converse true? 
i.e.

Given that $\frac{G}{H} \cong \frac{G}{K}$ then $H \cong K$ ?

I feel that the answer is no, but I can't think of an example.

REPLY [18 votes]: Let $$G = \mathbb Z/4\mathbb Z\times\mathbb Z/2\mathbb Z$$
and consider the subgroups
$$H = \mathbb Z/4\mathbb Z\times \{e\}\\K=\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z$$
Then $$G/K\cong G/H\cong\mathbb Z/2\mathbb Z$$ but $H\not\cong K$.

REPLY [6 votes]: Take, for instance, $G=\mathbb Z/4\mathbb Z \times \mathbb Z/2\mathbb Z$, $H=\mathbb Z/4 \mathbb Z \times \mathbf 0$, and $K=\mathbb Z /2\mathbb Z \times \mathbb Z/2 \mathbb Z$, so that $G/K\cong G/H \cong \mathbb Z/2 \mathbb Z$.